Find the differentiation of given algebraic equation \[\dfrac{dy}{dx}\]if \[{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}\]

Hint: We know that on differentiation of ${x}^{n}$ power of x becomes less than 1 and the derivative of constant is to be zero. So,here we use the derivative formulas of ${x}^{n} $ and constant.

Complete step-by-step solution -
We are given \[{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}\]
The equation can be re-written as \[{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}-{{a}^{\dfrac{2}{3}}}=0\]
Clearly , it is an implicit function . Hence , we can represent the function as \[f(x,y)\equiv {{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}-{{a}^{\dfrac{2}{3}}}=0\]
Now , we will differentiate the function \[f(x,y)\]with respect to \[x\].
 \[f(x,y)\] is an implicit function. So , to differentiate it, we will apply the chain rule of differentiation . According to the chain rule of differentiation , if \[R(x,y)\] is an implicit function, then the derivative of \[R(x,y)\] with respect to \[x\] is given by
\[\dfrac{dR}{dx}=\dfrac{\partial R}{\partial x}+\dfrac{\partial R}{\partial y}\times \dfrac{dy}{dx}\]
So, on differentiating both sides with respect to \[x\], we get ,
\[\dfrac{2}{3}.{{x}^{\left( \dfrac{2}{3}-1 \right)}}+\dfrac{2}{3}.{{y}^{\left( \dfrac{2}{3}-1 \right)}}.\dfrac{dy}{dx}=0\]
\[\dfrac{2}{3}{{x}^{\left( -\dfrac{1}{3} \right)}}+\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}.\dfrac{dy}{dx}=0\]
Now , we need to find the value of \[\dfrac{dy}{dx}\] . So , we will keep the terms with \[\dfrac{dy}{dx}\] on one side and all other terms on the other side.
On keeping terms with \[\dfrac{dy}{dx}\] on one side and all other terms on the other side , we get
\[\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}.\dfrac{dy}{dx}=-\dfrac{2}{3}{{x}^{\left( -\dfrac{1}{3} \right)}}\]
Now, we will shift \[\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}\] from the left-hand side to the right-hand side.
On shifting \[\dfrac{2}{3}{{y}^{\left( -\dfrac{1}{3} \right)}}\] from the left-hand side to the right-hand side , we get ,
\[\dfrac{dy}{dx}=\dfrac{-{{x}^{\dfrac{-1}{3}}}}{{{y}^{\dfrac{-1}{3}}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{x}{y} \right)}^{\dfrac{-1}{3}}}\]
Now, we will remove the minus sign from the exponent.
\[\Rightarrow \dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\]
Hence , the value of the derivative of the function \[{{x}^{\dfrac{2}{3}}}+{{y}^{\dfrac{2}{3}}}={{a}^{\dfrac{2}{3}}}\] is given by \[\dfrac{dy}{dx}=-{{\left( \dfrac{y}{x} \right)}^{\dfrac{1}{3}}}\].

Note: In the question , \[{{a}^{\dfrac{2}{3}}}\] is an arbitrary constant. So , on differentiating it with respect to \[x\], we get \[\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)=0\]
Some students consider it as a function of \[x\] and write \[\dfrac{d}{dx}\left( {{a}^{\dfrac{2}{3}}} \right)=\dfrac{2}{3}{{a}^{\dfrac{-1}{3}}}\dfrac{da}{dx}\]
Such mistakes should be avoided . Such mistakes will create confusion while solving the problem as the student will not be able to find the value of \[\dfrac{da}{dx}\].